A) \[\frac{\sqrt{3}+1}{2}\]
B) \[\frac{\sqrt{3}-\sqrt{2}}{2}\]
C) \[\frac{\sqrt{2}-1}{2}\]
D) \[\frac{\sqrt{3}-1}{2}\]
Correct Answer: D
Solution :
[d]| \[18{{x}^{2}}-9\pi x+{{\pi }^{2}}=0\] |
| \[(3x-\pi )(6x-\pi )=0\] |
| \[\Rightarrow \]\[\alpha =\frac{\pi }{6},\beta =\frac{\pi }{3}\] |
| \[f(x)=\sqrt{x},g(x)=\cos {{x}^{2}}\] \[(gof)(x)=\cos x\] |
| Area between \[\cos x,x=\frac{\pi }{6},x=\frac{\pi }{3}\] and \[y=0\] is \[A=\int_{\pi /6}^{\pi /3}{\cos xdx=[\sin x]_{\pi /6}^{\pi /3}}\] |
| \[A=\sin \frac{\pi }{3}-\sin \frac{\pi }{6}=\frac{\sqrt{3}-1}{2}\] |
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